For computing the fix field of $<\alpha \to -\alpha>$ we need to solve the equation $m+n\alpha+n'\alpha^2+n''\alpha^3=m-n\alpha+n'\alpha^2-n''\alpha^3$. Using the fact that $\{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\mathbb{Z}(\alpha)$ we have $n=n''=0$

For computing the fix field of $<\alpha \to \frac{1}{\alpha}>$ we need to solve the equation $m+n\alpha+n'\alpha^2+n'' \alpha^3=m+\frac{n}{\alpha}+\frac{n'}{\alpha^2}+ \frac{n''}{\alpha^3}$ samely since $\{1,\alpha,\alpha^2,\alpha^3\}$ is a base for $\mathbb{Z}(\alpha)$, we have $m\alpha^3+n+n'\alpha+n''\alpha^2=m\alpha^3+n \alpha^2+n ' \alpha + n''$
so $n=n''$

Galois group $\displaystyle \text{Gal}(L/\mathbb{Q}) $ of $\displaystyle L=\mathbb{Q}(\alpha) $ can be expressed as the group of following functions:
$\displaystyle
\varphi_1(x)=x,\ \varphi_2(x)=-x,\ \varphi_3(x)=\frac{1}{x},\ \varphi_4(x)=-\frac{1}{x}.
$

Using the fact that $\displaystyle \{1,\alpha,\alpha^2,\alpha^3\} $ is a base for $\displaystyle \mathbb{Q}(\alpha) $ (a basis of the linear space $\displaystyle \mathbb{Q}(\alpha) $) we have
$\displaystyle n=n''=0 $.